Ex. 7. Required the greatest common divisor of (d2-c3) Xa2+c1-d2c2 and 4da2-(2c2+4cd)a+ 202. Arranging these quantities according to d, we have (a2 c2)d2+c1—a2c2, or (a2 —c2)d2 — (a2 —c2)c2, and (4a2-4ac) xd-(a–c) x2c2; it is evident, by inspection only, that ac2 is a divisor of the first, and a-c of the second. But a2-c2 is divisible by a-c; therefore a-c is a divisor of the two proposed quantities: Dividing both the one and the other by a-c, the quotients will be (a+c)× (d2 — c2), and 4ad-2c2; which, by inspection, are found to have no common divisor, consequently a-c is the greatest common divisor of the proposed quantities. Ex. 8. Required the greatest common divisor of yx and y-y2x-yx2+x3. 3 2 2 Ans. y2-x Ex. 9. Required the greatest common divisor of a-b1 and a®-b®. Ans. a2-b2. Ex. 10. Required the greatest common divisor of a1+a3b-ab3—ba and aa +a2b2+ba. Ans. a+ab+b2, Ex. 11. Required the greatest common divisor of a2-2ax+x2 and a3-a2x-ax2+x3. Ans. a-x. Ex. 12. Find the greatest common divisor of 6x3-3x2+2y2x and 12x2-15yx+3y2. -- Ans. x-y. Ex. 13. Find the greatest common divisor of 36b2a6-18b2a5 — 27b2 a1 +9b2a3 and 2762a51862a4-9b2a3. Ans. 962a-9b2a3. Ex. 14. Find the greatest common divisor of (c-d)a2+(2bc-2bd)a+(b3c-b2d) and (bc-bd+ c2-cd)a+(b2d+bc2-b2c-bcd). Ans. c- -d. 2 Ex. 15. Find the greatest common divisor of qnp3+3np3 q2 —2npq3-2nq* and 2mp2 q2-4mp1 -mp3q+3mpg3. Ans. q-p. Ex. 16. Find the greatest common divisor of x3+9x2+27x-98 and x2+12x-28. Ans.x-2. § III. METHOD of finding the LEAST COMMON MUL TIPLE OF TWO OR MORE QUANTITIES. 145. The least common multiple of two or more quantities is the least quantity in which each of them is contained without a remainder. Thus, 20abc is the least common multiple of 5a, 4ac, and 26. 146. The least common multiple of two numbers, or quantities, is equal to their product divided by their greatest common measure, or divisor. For, let a and b be any two quantities, whose greatest common measure, or divisor, is x, and let a=mx, b=nx; then mnx is a multiple of a by the units in n, and of b by the units in m; consequently it is a common multiple of a and b. But since x is the greatest common measure of a and b, m and n can have no common divisor; mnx is, therefore, the least common multiple of a and b. Now mx=a, and nx=b; therefore mx Xnx=a×b, ab (Art. 50), and mnx=-, (Art. 51). Hence the rule x is evident; as for example: Let the least common multiple of 18 and 12 be required. Their greatest common measure is 6; therefore their least common multiple is. 12 X 18 6 -36. Every other common multiple of a and b is a multiple of mnx. Let q be any other common multiple of the two quantities, and, if possible, let mnx be con tained in q, times, with a remainders, which is less than mnx; then q―rmnx=s; and since a and b measure q and rmnx, they measure q-rmnx, or s, (Art. 131); that is, they have a common multiple less than mnx, which is contrary to the supposition. To find the least common multiple of three quantities a, b, and c. Let m be the least common multiple of a and b, and n the least common multiple of mand c; then is the least common multiple sought. For every common multiple of a and b is a multiple of m, therefore every common multiple of a, b, and c, is a multiple of m and c; also, every multiple of m and c is a multiple of a, b, and c; consequently the least common multiple of m and c is the least common multiple of a, b, and c. Or, in general, let a, b, c, d, &c. be any set of quantities, and let x be the greatest common divisor of a and b ; y ab of and c, &c. be the least common multiplier of a and b : 147. Hence the following method for finding the least common multiple of two or more quantities. RULE. I. Find the least common multiple of two quantities, by the preceding Article. II. Find, in like manner, the least common multiple of the result, thus found, and the third quantity. III. Find also the least common multiple of the last result and the fourth quantity; and proceed in the same manner with this result and the fifth; and so on the result last found will be the least common multiple of all the quantities. Ex. 1. Required the least common multiple of a3b2x, acbx2, and abc2 d. Here, the greatest common measure of a3b2x and acbx2, is abx, and the least common multiple is, a3b2 x Xacb x2 -=a3b2cx2; the greatest therefore, abx common measure of a3b2cx2 and abcd is abc; · a3b2 cx2 Xabc2 d hence, abc tiple required. =a3 b2 c2 dx2 = the least mul Ex. 2. Required the least common multiple of 2a2x, 4ax2, and 6×3. Here, the greatest common measure of 2a x and 4ax2, is 2ax; hence, the least common multiple is 2a2 x X4αx2 2ax 4a2x2; again, the greatest common measure of 4a3x3 and 6x3 is 2x2; and therefore 3 4a2x2 X6x3 12a3 3 the least common multiple 2x2 required. *Ex. 3. Required the least common multiple of 3a, 4a2, and 12ab. Ans. 24a2b. Ex. 4. Required the least common multiple of a2-b2, a+b, and a2+b2. Ans. a-b4. Ex. 5. Required the least common multiple of 27a, 15b, 9ab, and 3a2. Ans. 135a2b. Ex. 6. Required the least common multiple of a3+3a2b+3ab2 +b3, a2+2ab+b2, a2 —b2. Ans. a42a3b-2ab3-b4. Ex. 7. Required the least common multiple of a+b, a−b, a2+ab+b2, and a2 —ab+b3. Ans. a-b2. § IV. REDUCTION OF ALGEBRAIC FRACTIONS. CASE I. To reduce a mixed quantity to an improper fraction. RULE. 148 Multiply the integral part by the denominator of the fraction, and to the product annex the numerator with its proper sign: under this sum place the former denominator, and the result is the improper fraction required. 2h Ex. 1. Reduce 3x+2 to an improper fraction. 5a The integral part 3x multiplied by the denominator 5a of the fraction plus the numerator (26) is equal to 3x X5a+26=15ax+2b: 15ax+26 Hence, is the fraction required. 5a |